So the problem statement goes like this –
Points A, B and C lie on a circle and are arranged in clockwise order. We roll a fair 6-sided die and perform any of the 3 following operations based on the roll of the die.
- Move Clockwise – if 1 or 2 shows up.
- Move Anti-Clockwise – if 3 or 4 shows up.
- Do not move – if 5 or 6 show up.
We take A as our starting point. We roll the dice 8 times and perform the operations based on the dice roll.
What is the probability that we end up getting back to A.
This is how A, B & C lie on the circle
I want to solve the problem considering valid cases out of the various possible $3^8$ cases.
I looked up the editorial for this and it stated the answer to be 1/3 due to symmetry (equal probabilities of performing ay of the operations.) However, I wanted to try out forming the valid cases by myself and was unable to reach the answer –
My approach :
Let us call
cc – total clockwise operations |
ac – total anticlockwise operations |
xx – total no movement operation
We can have
- Equal Number of Both Clockwise and Anticlockwise cases.
- If ac or cc is a multiple of 3 and the other is 0.
- If not a multiple of 3 and unequal, then ( (cc%3) – ac) || ((ac%3) – cc) == 0 ) where % stands for modulus whih yields the remainder
My valid cases – (ac, cc, xx)
- (1, 1, 6) = 56
- (2, 2, 4) = 840
- (3, 0, 5) = 56
- (5, 0, 3) = 56
- (3, 3, 2) = 560
- (4, 4, 0) = 70
- (4, 1, 3) = 280
- (1, 4, 3) = 280
- (5, 2, 1) = 168
- (2, 5, 1) = 168
- (6, 0, 2) = 28
- (0, 6, 2) = 28
- (7, 1, 0) = 8
- (1, 7, 0) = 8
- (0, 0, 8) = 1
Total equals = 2607.
When I divide it by $3^8$, I get it as 0.397 (not exactly 0.33).
Am I missing certain cases ?
Best Answer
You can't be (and don't seem to be) missing cases because you're getting too large an answer. The more likely error is that you're miscounting combinations that lead to a particular case. (You should have $(0, 3, 5)$ instead of $(5, 0, 3)$ but that won't affect your result.)
You should end up with $2187 (= 3^7)$ combinations that work. By my calculation, there are $420=\frac{8!}{4!2!2!}$ combinations that result in $(2, 2, 4)$, not $840$ as set forth in your calculation. I think that exactly accounts for your error.